Maximization problems

Question

A box with an open top is to be constructed from rectangular sheet of cardboard of size 5 feet by 8 feet. You are to cut out 4 corner squares of the cardboard in such a way that the volume of the box is as large as poossible. Let x denote the length in feet of the side of the square cutout.

A) write the volume of the box as a function of x.

b) what is the domain of the function in part a?

c) what is the derivative of the function in part a?

d) for what values of x is the derivative equal to 0?

Answer 1

I recommend that you always start by drawing a picture.

After you cut out the 4 squares, you will have a shape of a cross. Each arm of the cross will fold to form one of the four sides of the box. The rectangle in the center of the cross will be the bottom of the box.

The variable x is the side of each square cut out---it is also the length of each arm of the cross, and it therefore ALSO the height (H) of the completed box.

Next, recognize that the length and width of the box are as follows:

Length (L) = 8 - 2x

Width (W) = 5 - 2x

Now we can write the equation for the volume of the box:

V = L*W*H = (8-2x)*(5-2x)*x

Expand and simplify to get:

V = 4x³ - 26x² + 40x

Domain: This is the allowable x values. Mathematically, x can be zero, which yields a box with zero volume. The maximum for x occurs when the square cutouts consume the entire 5-foot dimension of the cardboard. At that point, x is 2.5, and the width of the box is zero.

So: 0 <= x <= 2.5

To find the maximum volume, take the 1st derivative of the equation, and set equal to 0:

V' = 12x² - 52x + 40 = 0

factor out the constant 4:

3x2 - 13x + 10 = 0

Factor to get (3x - 10)*(x - 1) = 0

Thus, there is a minimum or maximum at x = 10/3 and x = 1. I think you will find that 10/3 gives the maximum. (To verify, plug the values into the equation for V)